All categories

2 years ago

An equipoterntial surface that surrounds a + 3.0 pC point charge has a radius of 2.0 cm. What is the potential of this surface?

Physics

1 answer:

mestny [16]2 years ago

8 0

Answer:

Electric potential = 0.00054 V

Explanation:

We are given;

Charge; q = 3 pC = 3 × 10^(-12) C

Radius; r = 2 cm = 0.02 m

Formula for the electric potential of this surface will be;

V = kqr

Where;

K is a constant = 9 × 10^(9) N⋅m²/C².

Thus;

V = 9 × 10^(9) × 3 × 10^(-12) × 0.02

V = 0.00054 V

You might be interested in

A 2 microcoulomb charge is placed at a distance of 0.25 m away from a 3.6 microcoulomb charge. Describe the type of electrostati

EleoNora [17]

Answer: 1.04N

Explanation:

Given

q1 = 2*10^-6C

q2 = 3.6*10^-6C

r = 0.25m

k = 9*10^9

Magnitude of electrostatic force can be calculated by using coulomb's law. Coulomb's law states that, "the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them."

F =(kq1q2) / r²

F = (9*10^9 * 2*10^-6 * 3.6*10^-6) / 0.25²

F = 0.0648/0.0625

F = 1.04N

The type of electrostatic force between the charges is the repulsive force

7 0

2 years ago

Read 2 more answers

CAN SOMEONE PLEASE TELL ME WHAT THIS WHEEL IS CALLED.WILL GIVE BRAINLIEST

kolbaska11 [484]

Answer:

its a pie chart

Explanation:

.........................

3 0

3 years ago

Read 2 more answers

Find the intensity of the electromagnetic wave described in each case. (a) an electromagnetic wave with a wavelength of 630 nm a

qaws [65]

Find the intensity of the electromagnetic wave described in each case.

(a) an electromagnetic wave with a wavelength of 645 nm and a peak electric field magnitude of 8.5 V/m.

(b) an electromagnetic wave with an angular frequency of 6.3 ✕ 1018 rad/s and a peak magnetic field magnitude of 10−10 T.

3 0

3 years ago

In si units, the electric field in an electromagnetic wave is described by ey = 104 sin(1.40 107x − ωt). (a) find the amplitude

melamori03 [73]

Answers:
(a) $B_o = 0.3466$ μT
(b) $\lambda = 0.4488$ μm
(c) f = $6.68 * 10^{14}Hz$

Explanation:
Given electric field(in y direction) equation:
$E_y = 104sin(1.40 * 10^7 x -\omega t)$

(a) The amplitude of electric field is $E_o = 104$ . Hence

The amplitude of magnetic field oscillations is $B_o = \frac{E_o}{c}$
Where c = speed of light

Therefore,
$B_o = \frac{104}{3*10^8} = 0.3466$ μT (Where T is in seconds--signifies the oscillations)

(b) To find the wavelength use:
$\frac{2 \pi }{\lambda} = 1.40 * 10^7$
$\lambda = \frac{2 \pi}{1.40} * 10^{-7}$
$\lambda = 0.4488$ μm

(c) Since c = fλ
=> f = c/λ

Now plug-in the values
f = (3*10^8)/(0.4488*10^-6)
f = $6.68 * 10^{14}Hz$

6 0

2 years ago

For a certain RLC circuit the maximum generator EMF is 125 V and the maximum current is 3.20 A. If le a) the impedance of the ci

MAXImum [283]

Answer:

Part (i)

Z = 39.06 ohm

Part (ii)

R = 21.7 ohm

Explanation:

a) here we know that

maximum value of EMF = 125 V